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Mole Fraction

• The mole fraction is a unit of concentration, defined to be equal to the number of moles of a component divided by the total number of moles of a solution
• The mole fraction is determined for all reactants in a mixture of gases
• g. to find the mole fraction of A (X_a):

Partial pressure

• Partial pressure it the pressure exerted by the single gas if it were isolated
• The total pressure is the combined total of all the partial pressures
• To determine the partial pressure for one gas, multiply the mole fraction by the total pressure
• g. to find the partial pressure of A (P_A):
• Units for partial pressure Pascals or atmospheres

P_A = Mole Fraction of A  x Total Pressure

P_A = X_a  x   P

K_p in homogeneous gaseous equilibria

• • Homogeneous Equilibria refers to all involved reactants are in the same phase
  • For Kp to function, everything must be a gas

• Kp will always be a constant except for a change in temperature

K_p in heterogeneous equilibria

• Heterogeneous equilibrium will involve gases in contact with solids
• As Kp functions only with a gas, the solid is not included

K_c

• With a known molar concentration, of each substance at equilibrium then Kc can be determined
• Kc is only specific to that particular temperature

aA + bB ⇋ cC + dD

Kc = [C]^c [D]^d
[A]^a [B]^b

• ^­Lowercase letters represent the number of each substance
• Square brackets refer to the concentration (mol dm^-3)
• The products go on the tops and the reactants at the bottom

e.g. 2Mg + O₂  -> 2MgO
Kc = [MgO]² 
[Mg]² [O₂]

Method:

• Find the amount of moles of each reactant and product there are at equilibrium (find this by balancing the equation if not given)
• Calculate the molar concentration of each reactant:

Concentration (mol dm^-3) = Number of moles (mol)
Volume (dm^-3)

• Place these values into the Kc equation
• To determine the units: each value in the brackets is ‘mol dm-3’ and multiply it by the amount power and cross off the the units that are on both sides of the fraction e.g.

Kc = [MgO]² =  (mol dm^-3 ) (mol dm^-3)
[Mg]² [O₂] = (mol dm^-3) (mol dm^-3) (mol dm^-3)

Unit = mol dm^-3

1. If there are an equal number of ‘mol dm^-3’ then there is no units
2. If there are 2 or more mol dm^-3 then multiply it by how many there are e.g.

2 x mol dm^-3  = (mol^-2 dm^-6)

Using Kc

With the value of Kc known, then an unknown equilibriums concentration can be determined;

• Put all of the values you know into the expression of Kc
• Rearrange the equation and solve it for the unknown values

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Dynamic Equilibrium

• Many reactions are reversible
• Dynamic equilibrium is reached when the concentration of the reactants and products are at a constant, and the forward and backward reactions are at the same rate
• Dynamic equilibrium can only occur in a closed system at a constant temperature
• An example of dynamic equilibrium would be more reactants being used up in the reaction would cause the forward reaction to slow down as more products are forming. However as there are now more products than there are reactants the reaction will shift to the left producing more reactants. After a while there will be an equal quantity of both reactants and product forming and therefore the overall concentration of each does not change.

Le Chatelier’s principle

“if a system at equilibrium is disturbed, the equilibrium shifts in the direct that tends to reduce the disturbance”

• Any factor which changes the effect of the equilibrium mixture will result in the position of equilibrium shifting so to oppose the change
• Le Chatelier’s Principle does not account for how far the equilibrium changes, and therefore no quantities can be predicted

Change in Concentration

• If an increase in concentration of one of the reactants then the equilibrium will shift in the direction that will reduce the concentration

A_(aq) + B_(aq)⇋ C_(aq) + D_(aq)

• by increasing the concentration of B this will react with A which will produce more C+D. Thus, more product is formed and equilibrium moves to the right.
• By removing a product formed, such as C, then the equilibrium will move to the right to produce more C (as well as D) using up the reactants A+B more.

Changing the Pressure

• Changing the pressure only affects reaction that include gases
• Changing the pressure only affects gas reactions that have a differing number of molecules on each side

A_(g)⇋ 2C_(g)

• Increasing the pressure of a gas means that there are more molecules in it at a given volume, it is equivalent to increasing the concentration of a solution
• Increasing the pressure of a system will cause the position of equilibrium to shift to the side with the fewer molecules as that will reduce the pressure

Changing the temperature

• All reversible reactions have an exothermic reactions in one direction and an endothermic reaction in the opposite
• A negative ΔH = exothermic
• A positive ΔH = endothermic

A_(g) + B_(g)⇋ C_(g) + D_(g) ΔH = -203 Kj mol^-1

• As the ΔH is negative, the forward reaction is exothermic (reactants -> products)
• The backwards reaction is endothermic (products -> reactants)
• By increasing the temperature, the equilibrium will move to the endothermic reaction. As endothermic reactions absorb heat
• By decreasing the temperature, the equilibrium will move to the exothermic reaction. As exothermic reactions give off heat

Catalysts

• Catalysts do not affect the position of equilibria
• They allow for equilibrium to be reached more quickly as they effect both side of the reaction equally